Someone please explain the Equilibrium Model Pract (1 Viewer)

123ryoma12 · Oct 4, 2016

In particular the one with two different sized pipettes.
In was in the 2011 HSC sample answer.

"Two identical measuring cylinders are filled with different volumes of water. Water is
transferred backwards and forwards from each cylinder using two differently sized pipettes
until the volume of water in each cylinder remains constant (but at different levels). "

According to many people's notes, they fill Cylinder A with 100 mL water and leave Cylinder B empty.
Then they use a 10 mL pipette to transfer water from Cylinder A to Cylinder B. Then a smaller pipette (say, 5 mL) to transfer water from Cylinder B to Cylinder A.
But if you continue to do this, wouldn't you run out of water in Cylinder A?
Because the rate at which your losing water in cylinder A and gaining water in cylinder B is constant, unless the size of the pipette changes.

InteGrand · Oct 5, 2016

123ryoma12 said:
In particular the one with two different sized pipettes.
In was in the 2011 HSC sample answer.

"Two identical measuring cylinders are filled with different volumes of water. Water is
transferred backwards and forwards from each cylinder using two differently sized pipettes
until the volume of water in each cylinder remains constant (but at different levels). "

According to many people's notes, they fill Cylinder A with 100 mL water and leave Cylinder B empty.
Then they use a 10 mL pipette to transfer water from Cylinder A to Cylinder B. Then a smaller pipette (say, 5 mL) to transfer water from Cylinder B to Cylinder A.
But if you continue to do this, wouldn't you run out of water in Cylinder A?
Because the rate at which your losing water in cylinder A and gaining water in cylinder B is constant, unless the size of the pipette changes.

$\noindent You're right in that if the pipettes did something like always move 10 mL and 5 mL back and forth, we wouldn't reach an equilibrium (well we'd get A end up empty and B end up full and). So I'm going to assume the following is how the transferring works: if a beaker has height $h$ of water in it, then a pipette of radius $r$ (and length bigger than $h$) will transfer a volume of $\pi r^2 h$ (i.e. the pipette draws up a cylinder of water with height whatever the depth of water in the beaker was. The volume of this cylinder is $\pi r^{2}h$, since $r$ is the radius of the pipette (and hence the cylinder of water)). Doing this with heights, we can show an equilibrium will be reached. We show this below. (This is a pretty good approximation to reality.)$

$\noindent Assume the pipettes are cylinders and beakers are identical cylinders. Let the base area of the beakers be $z$. Assume the pipettes have a longer length than the beakers' heights, so it can always transfer a full amount if necessary. Let the smaller pipette have radius $r$ and the bigger one have radius $R > r$. Let beakers have height $H$ and beaker A start off full, B empty. Call transferring from A to B and then from B to A a cycle. Let $A_n$ be the HEIGHT of water in cylinder $A$ after the $n$-th cycle. Let $B_n$ be the that for cylinder $B$. We have that $A_0 = H$ is the initial height of water in beaker $A$ and $B_0 = 0$ the initial height of water in B. Note by conservation of water, $A_n + B_n = H$ for all $n \in \mathbb{N}$. The volume of water in beaker $A$ after the $n$th cycle is $zA_n$ (base area times height).$

$\noindent Now, what happens to A during the $n$th cycle? We transfer water from A to B, then back from B to A. Just before the $n$th cycle, height of water in A is $A_{n-1}$ and height in B is $B_{n-1} = H - A_{n-1}$. The volumes therefore are $zA_{n-1}$ and $zH -zA_{n-1}$ respectively for A and B. When we lower the first (bigger) pipette into A now, we withdraw a volume of $\pi R^2 A_{n-1}$, by assumption. So we put this into B, making A's volume now $z A_{n-1} - \pi R^2 A_{n-1} = \left(z -\pi R^2\right)A_{n-1}$. The height of A at this stage is its volume divided by base area $z$, i.e. $\left(1 -\frac{\pi R^2}{z}\right)A_{n-1}$. So the height of water in B right now is $H - \left(1 -\frac{\pi R^2}{z}\right)A_{n-1}$.$

$\noindent We now transfer water from B to A using the pipette of radius $r$. How much water is transferred? A volume of $\pi r^2 H - \pi r^2 \left(1 -\frac{\pi R^2}{z}\right)A_{n-1}$ (i.e. $\pi r^2$ times height of water in B). So at the end of the $n$th cycle, volume of water in A has become$

$$$\begin{align*}\left(z -\pi R^2\right)A_{n-1} + \pi r^2 H - \pi r^2 \left(1 - \frac{\pi R^2 }{z}\right)A_{n-1} &= \pi r^{2} H + \left(z -\pi R^2 - \pi r^{2} + \frac{\pi^{2} r^{2}R^{2}}{z}\right)A_{n-1}.$$\end{align*}$

$\noindent Divide through by $z$ to get the height now, which is $A_n$. Hence $A_n = C + D A_{n-1}$, where $C = \frac{\pi r^2 H}{z}$ and $D = \frac{z -\pi R^2 -\pi r^2 + \frac{\pi ^2 r^2 R^2}{z}}{z} = \frac{z^{2} - \pi \left(R^{2} +r^{2}\right)z + \pi^{2}r^{2}R^{2}}{z^{2}}$.$

$\noindent So we have the recurrence $A_{n} = C + DA_{n-1}$. So $A_{n} = C + D(C + DA_{n-2}) = C + CD + D^{2}A_{n-2}$. If we continue the pattern, we'll find$

$\begin{align*}A_{n} &= C + CD + \cdots + CD^{n-1} + D^{n} A_{0}\\ &= C + CD + \cdots + CD^{n-1} + HD^{n}.\end{align*}$

$\noindent This is good! We have an expression for $A_{n}$ directly in terms of $n$ now, and we can simplify this using geometric series formulas. But first we need to see what $D$ is like. This is because the long-term behaviour of $A_{n}$ will depend on this.$

$\noindent Recall that $D = \frac{z^{2} - \pi \left(R^{2} +r^{2}\right)z + \pi^{2}r^{2}R^{2}}{z^{2}} = 1 - \frac{\pi \left(R^{2} +r^{2}\right)}{z} + \frac{\pi^{2}r^{2}R^{2}}{z^{2}}$ (in real life, $z > 0$). When could $D$ be $0$? This happens iff $z^{2} - \pi \left(R^{2} +r^{2}\right)z + \pi^{2}r^{2}R^{2} = 0$. The discriminant is $\Delta = \pi^{2} \left(R^{2} +r^{2}\right)^{2} - 4\pi^{2} r^{2}R^{2} = \pi^{2} \left(R^{2}-r^{2}\right)^{2}$. By the quadratic formula then, $D = 0$ precisely when$

$\begin{align*}z &= \frac{ \pi \left(R^{2} +r^{2}\right)\pm \pi \left(R^{2}-r^{2}\right)}{2}\\ &= \pi r^{2} \text{ or }\pi R^{2}.\end{align*}$

$\noindent The numerator of $D$ is a quadratic in $z$, so we see that $D > 0$ for $ z > \pi R^{2}$ or $z < \pi r^{2}$ (since $R > r$). In practice, $z$ is going to be much greater than $\pi R^{2}$ in fact (i.e. the base of the beaker is much bigger than the cross-sectional area of the larger pipette). So in practice, we always have $D > 0$.$

$\noindent Now, we claim that in practice we also have $D < 1$. To see this, we can sketch the graph of $D$ vs. $z$. Note that since $D = 1 - \frac{\pi \left(R^{2} +r^{2}\right)}{z} + \frac{\pi^{2}r^{2}R^{2}}{z^{2}}$, $D \to + \infty$ as $z \to 0^{+}$, since there is a positive coefficient of $\frac{1}{z^{2}}$ here, and this dominates any impact of the negative contribution from $\frac{1}{z}$ as $z\to 0^{+}$. Also, we already know the $z$-zeros of $D$ are at $z = \pi r^{2}$ and $\pi R^{2}$. Moreover, note that $\frac{\partial D}{\partial z} = \frac{\pi \left(R^{2} + r^{2}\right)}{z^2} - \frac{2\pi^{2} r^{2} R^{2}}{z^{3}}$, which is $0$ when $z = z_{0} = \frac{2\pi r^{2}R^{2}}{R^{2} + r^{2}}$. Note that $z_{0}$ is between $\pi r^{2}$ and $\pi R^{2}$, which follows from the uniqueness of the stationary point and Rolle's Theorem (but can also easily be proved algebraically).$

$\noindent So $D$ viewed as a function of $z$ has a stationary point in between its two roots. Also, $D\to 1^{-}$ as $z\to \infty$, and since we noted $D \to +\infty$ as $z\to 0^{+}$, the stationary point is a local minimum. Since $D$ had a unique stationary point wrt $z$, $D$ will be increasing after $z = z_{0}$, approaching $1$ as $z \to \infty$. In particular, since $D = 0$ when $z = \pi R^{2}$, we have $0 < D < 1$ for all $ z > \pi R^{2}$, and the graph of $D$ vs. $z$ looks something like this:$

http://www.graphsketch.com/?eqn1_co..._lines=1&line_width=4&image_w=850&image_h=525 .

$\noindent So in practice ($z > \pi R^{2}$), we have $0 < D < 1$. This is good too -- it completely explains the equilibrium. Because now,$

$\begin{align*}A_{n} &= C + CD + \dots + C D^{n-1} + HD^{n} \\ &= C \times \frac{1 - D^{n}}{1-D} + HD^{n} \quad (\text{where }D \text{ is a constant between }0 \text{ and }1).\end{align*}$

$\noindent Since $0 < D< 1$, as $n\to \infty$, $D^{n} \to 0$, and so we see that we indeed have an equilibrium in the height in beaker A, with $A_{n} \to \frac{C}{1-D}$ as $n \to \infty$, where $C = \frac{\pi r^2 H}{z}$ and $D = \frac{z -\pi R^2 -\pi r^2 + \frac{\pi ^2 r^2 R^2}{z}}{z}$.$

$\noindent So in summary, due to the assumption about how transferring water works, an equilibrium height in beaker A is reached, and this height is given above as $\frac{C}{1-D}$, which you could calculate if you could measure the base area and height of your beaker and the squares of the radii of your pipettes (you could do the latter approximately by finding the volume of water the pipettes can hold, and measuring the height $h$ of the pipettes, then the squared radius of the pipette is the volume divided by $\pi h$).$

Someone please explain the Equilibrium Model Pract (1 Viewer)

123ryoma12

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InteGrand

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